Are the harmonic maps from a 2-dimensional cone to $S^2$ locally lipschitz or Holder continuous?
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They are localy Holder continuous. There is a map $\phi$ from the plane to the cone which is bijective, the inverse is Holder, and $\phi$ is conformal except one point. This map can be written explicitly. Let your map be $f$. Composition $g=f\circ\phi$ is harmonic except one point, so by removable singularity theorem it is harmonic everywhere, thus locally Holder. So your map $f=g\circ\phi^{-1}$ is locally Holder. The Holder exponent depends on the opening of the cone.
answered Feb 17, 2014 at 14:53
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$\begingroup$ :Just as you have said before. Take the cone as a sector with total angel $\alpha<2\pi$. Then $\phi^{-1}$ is $z\mapsto z^{\pi/\alpha}$. It seems to me this map is also locally Lipshcitz(hence every harmonic map is locally lipschitz). $\endgroup$ Commented Feb 18, 2014 at 13:13
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$\begingroup$ In your question you did not specify the range of $\alpha$. In general, this map is Holder. $\endgroup$ Commented Feb 18, 2014 at 14:16